Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $q = \dfrac{r + 7}{-2r - 2} \times \dfrac{r^2 - 1}{-6r + 6} $
Solution: First factor the quadratic. $q = \dfrac{r + 7}{-2r - 2} \times \dfrac{(r + 1)(r - 1)}{-6r + 6} $ Then factor out any other terms. $q = \dfrac{r + 7}{-2(r + 1)} \times \dfrac{(r + 1)(r - 1)}{-6(r - 1)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (r + 7) \times (r + 1)(r - 1) } { -2(r + 1) \times -6(r - 1) } $ $q = \dfrac{ (r + 7)(r + 1)(r - 1)}{ 12(r + 1)(r - 1)} $ Notice that $(r - 1)$ and $(r + 1)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ (r + 7)\cancel{(r + 1)}(r - 1)}{ 12\cancel{(r + 1)}(r - 1)} $ We are dividing by $r + 1$ , so $r + 1 \neq 0$ Therefore, $r \neq -1$ $q = \dfrac{ (r + 7)\cancel{(r + 1)}\cancel{(r - 1)}}{ 12\cancel{(r + 1)}\cancel{(r - 1)}} $ We are dividing by $r - 1$ , so $r - 1 \neq 0$ Therefore, $r \neq 1$ $q = \dfrac{r + 7}{12} ; \space r \neq -1 ; \space r \neq 1 $